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How do we calculate the Cartesian product between $n$ polygons? Is it always a polytope? If so, can we say anything about its faces?

For example, $n$ squares: $\{S_1:[a_1, a_2] \times [b_1, b_2], \ldots, S_n:[a_k, a_{k+1}] \times [b_{k}, b_{k+1}]\}$ so, $S_1 \times \ldots \times S_n \in \mathcal{R}^n$ forms a $n-$dimensional box.

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    Do you mean, what is the resulting polytope? – 2017-02-18
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    Yes. For example: the cartesian product between $n$ hexagon. What form does this product have? Is it a polytope? How? – 2017-02-18
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    OK, I tried to clean up the question a bit -- I hope I haven't put too many words into your mouth (feel free to keep editing). I only did this because the question currently has 3 out of 5 votes to close, and I'd rather it not be closed. Anyway, yes, it's always a polytope. The product of $n$ squares is actually $(2n)$-dimensional (so in $\Bbb R^{2n}$). If I find time in the next couple days, I'll try my hand at an answer if the question remains open. – 2017-02-18

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Cartesian products of polytopes $P,Q$ are indeed polytopes again. The vertices are pairs of vertices $(p,q)$ where $p$ is a vertex of $P$, and $q$ is a vertex of $Q$.

In fact, if you have two polytopes, say $P \subset \mathbb{R}^n$ and $Q \subset \mathbb{R}^m$, then all faces of $P \times Q$ are just the Cartesian products of faces of $P$ and $Q$. You can see this by using the hyperplane descriptions: the hyperplanes used to define $P \times Q$ are the same as those for $P$ and $Q$, but now they are using disjoint sets of variables. So, if $F \subset P \times Q$ is a face, then the first $n$ coordinates of every point in $F$ must satisfy equalities for some face of $P$ and the last $m$ coordinates of every point in $F$ must satisfy equalities for some face of $Q$.

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Just to add some relevant terms. The cartesian product of polytopes also is known as the prism product.

In fact, a polytope $P$ multiplied by some single vertex point $v$ results in $P\times v$, which is $P$ again (Identity). Multiplication by a single edge $e$ results in $P\times e$, which is the usual prism with bases being $P$ and lacing edges $e$. If the dimension of $Q$ also is greater than 1, the product $P\times Q$ is known to be the ($P,Q$)-duoprism, cf. https://en.wikipedia.org/wiki/Duoprism, and using even more factors, you'll get multiprisms (like ($P,Q,R$)-triprisms $P\times Q\times R$, then ($P,Q,R,S$)-quadprisms $P\times Q\times R\times S$, etc.)

The total dimension is calculated according to
$$d({\huge\times}P_k)=\sum{d(P_k)}$$ And, as @RobDavis already said, this applies even for any elements thereof. Thus the vertices of ${\LARGE\times}P_k$ are given by the respective products ${\LARGE\times}v_k$, the cartesian products of the vertices each. In order to result in an edge of ${\LARGE\times}P_k$, all components have to be vertices, just a single factor is an edge within the respective polytopal subspace. Similarily 2-faces of ${\LARGE\times}P_k$ occure from all vertices and one polygonal factor, or from all vertices and 2 factors being edges, which then results in the correspondingly oriented rectangle.

For even more details on this prism product, and on even other polytopal products, cf. my webpage at https://bendwavy.org/klitzing/explain/product.htm.

--- rk